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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Kinetic Theory of Gases

BITSAT 2024 · Paper 2 · Physics · Kinetic Theory of Gases

The temperature of an ideal gas is increased from 200 K to 800 K....

The temperature of an ideal gas is increased from 200 K to 800 K. If r.m.s. speed of gas at 200 K is v0v_0. Then, r.m.s. speed of the gas at 800 K will be:

Options

  1. Av0\mathrm{v}_0
  2. B4v04 v_0
  3. Cv04\frac{v_0}{4}
  4. D2v02 \mathrm{v}_0

Correct answer

D. 2v02 \mathrm{v}_0

Step-by-step solution

RMS speed, Vrms=3RTM⇒Vmm∝T\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}\Rightarrow \quad \mathrm{V}_{\mathrm{mm}} \propto \sqrt{\mathrm{T}} Here, Tinitiol =200KTfing1=800K\mathrm{T}_{\text{initiol }}=200 \mathrm{ K}\mathrm{T}_{\mathrm{fing1}}=800 \mathrm{ K} Initial RMS speed =v0∴v0vrms=200800⇒vms=2v0=\mathrm{v}_0\therefore \frac{v_0}{v_{\mathrm{rms}}}=\sqrt{\frac{200}{800}} \Rightarrow v_{\mathrm{ms}}=2 v_0
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